How does complexifying a Lie algebra $\mathfrak{g}$ to $\mathfrak{g}_\mathbb{C}$ help me discover representations of $\mathfrak{g}$?

Question

I have been studying a course on Lie algebras in particle physics and I could never understand how complexifying helps us understand the original Lie algebra.

For example, consider $\mathfrak{su}(2)$: I complexify this to give me $\mathfrak{su}(2)_\mathbb{C}$ which allows me to form a Cartan-Weyl basis of ladder operators and a Cartan subalgebra, so I can generate highest weight representations. This is fine, but these are representations of $\mathfrak{su}(2)_\mathbb{C}$ not $\mathfrak{su}(2)$ because we cannot form a Cartan-Weyl basis unless we complexify. So how does this help us construct representations of $\mathfrak{su}(2)$?

All across particle physics we talk about particles living in representations of particular Lie algebras, but in fact should they actually be living in the complexifications of these? i.e. we talk about spinors of the Lorentz group, but the way to come across these is by complexifying the Lie algebra $\mathfrak{so}(3,1)$ so that it decomposes as

$$\mathfrak{so}(3,1)_\mathbb{C}=\mathfrak{su}(2)_\mathbb{C} \oplus \mathfrak{su}(2)_\mathbb{C}.$$

At which point I can label representations by $(A,B)$ where $A,B$ label the highest weights of the two subalgebras, and I would say that left-handed spinors live in $(1/2,0)$ and right handed in $(0,1/2)$. But again, these are complexified Lie algebras. How does this tell me that spinors exist with respect to the real Lorentz group, the one which the universe uses.

In summary, my questions:

1. How does complexifying a Lie algebra $\mathfrak{g}$ to $\mathfrak{g}_\mathbb{C}$ help me discover representations of $\mathfrak{g}$ when the highest weight method only works with complexified Lie algebras?

2. How do I know that the things which I discover after complexifying, like spinors and particle multiplets, are valid with respect to the orginal Lie algebra? i.e. We live in a universe where Lorentz transformations are real NOT complex, so how can we discover spinors without complexifying our Lorentz algebra?

Answer 1

You just need to find what is the reality condition you should impose in your objects. For example, the

$$ SO(4)_\mathbb{C}=SL(2,\mathbb{C})\times SL(2,\mathbb{C}) $$

allows you to write a chiral spinor as $\chi_{\alpha}$ and an antichiral as $\chi_{\dot\alpha}$, each being in the fundamental representation of one of the $SL(2,\mathbb{C})$'s. Chirality does not change if you change the signature. The Lorentz generators are given by $M_{\alpha}^{\beta}$ and $M_{\dot\alpha}^{\dot\beta}$ and they are both traceless.

Now, if you want $SO(4)_{\mathbb{R}}=SU(2)\times SU(2)$ you should impose the reality condition

$$ (M^\alpha_\beta)^*=M_\alpha^\beta,\qquad, (\chi_\alpha)^*=\bar\chi^\alpha,\qquad (\chi_{\dot\alpha})^*=\bar\chi^{\dot\alpha}\qquad (\sigma^m_{\alpha\dot\alpha})^*=(\bar\sigma^m)^{\dot\alpha\alpha}\equiv\varepsilon^{\alpha\beta}\varepsilon^{\dot\alpha\dot\beta}\sigma^{m}_{\beta\dot\beta} $$

If you want $SO(3,1)_{\mathbb{R}}=SL(2,\mathbb{C})$ the relatity conditions are modified to:

$$ (M^\alpha_\beta)^*=M^{\dot\alpha}_{\dot\beta},\qquad (\chi_\alpha)^*=\chi_{\dot \alpha},\qquad (\sigma^m_{\alpha\dot\beta})^*=\sigma^m_{\beta\dot\alpha} $$

Finally the $SO(2,2)=SL(2,\mathbb{R})\times SL(2,\mathbb{R})$ you should impose:

$$ (M^\alpha_\beta)^*=M^\alpha_\beta,\qquad, (\chi_\alpha)^*=\bar\chi_\alpha,\qquad (\chi_{\dot\alpha})^*=\bar\chi_{\dot\alpha}\qquad (\sigma^m_{\alpha\dot\beta})^*=(\sigma^m)_{\alpha\dot\beta} $$

Note that this different reality conditions select different sigma matrices, that could be related by

$$ (\sigma^{4})_{SO(4)}=i(\sigma^{0})_{SO(3,1)},\quad (\sigma^{1,2,3})_{SO(4)}=(\sigma^{1,2,3})_{SO(3,1)} $$

$$ (\sigma^{0,2})_{SO(4)}=i(\sigma^{0,-1})_{SO(2,2)},\quad (\sigma^{1,3})_{SO(4)}= (\sigma^{1,3})_{SO(2,2)} $$

if you use the convention where

$$ P_{SO(4)}\equiv p_{m}(\sigma^{m})_{SO(4)} = \begin{bmatrix} p_3+ip_{4} & p_1-ip_{2}\\p_1+ip_{2} & -p_3+ip_{4} \end{bmatrix} $$

Note that

$$ \delta^{mn}_{SO(4)}p_{m}p_{n}= -\det(P_{SO(4)}) $$

which confirms that $SO(4)=SU(2)\times SU(2)$ in terms of Lie Algebra. The same follows for others signatures.

Answer 2

The main answer to OP's title question is that there is a bijective correspondence between representations for a real Lie algebra $\mathfrak{g}$ and its complexification $\mathfrak{g}_{\mathbb{C}}$ in the following sense:

1. Any representation of a Lie group/Lie algebra forms trivially a restricted representation wrt. to a Lie subgroup/algebra, respectively. Also, any $\mathbb{C}$-linear Lie algebra representation is trivially an $\mathbb{R}$-linear Lie algebra representation. It is the other direction that is interesting$^1$.

2. It is straightforward to complexify a real Lie algebra $\mathfrak{g}$ and its Lie bracket $$[x_1+iy_1,x_2+iy_2]_{\mathbb{C}}~:=~[x_1,x_2]-[y_1,y_2] + i[x_1,y_2]+i[y_1,x_2].$$

3. An $\mathbb{R}$-linear representation $\rho:\mathfrak{g}\to {\rm End}(V)$ of a real Lie algebra $\mathfrak{g}$ can be extended to the complexification $\mathfrak{g}_{\mathbb{C}}$ in the following way:

   • If $V$ is a $\color{red}{\mathbb{R}}$-vector space, replace $V$ with its complexification $V_{\mathbb{C}}$. The representation becomes an $\mathbb{R}$-linear representation $\rho:\mathfrak{g}\to {\rm End}(V_{\mathbb{C}})$ in the obvious way $$\rho(x)(u+iv)~:=~\rho(x)u+i\rho(x)v, \qquad u,v~\in~V.$$ We will rename $V_{\mathbb{C}}$ as $V$ from now on.

   • If $V$ is a $\color{red}{\mathbb{H}}$-vector space, pick a distinguished complex unit, say $i$, so that $V$ can be viewed as a $\mathbb{C}$-vector space.

   • If $V$ is a $\color{red}{\mathbb{C}}$-vector space, define the extension to a $\mathbb{C}$-linear representation $\rho_{\mathbb{C}}:\mathfrak{g}_{\mathbb{C}}\to {\rm End}(V)$ in the obvious way: $$\rho_{\mathbb{C}}(x+iy)~:=~\rho(x)+i\rho(y), \qquad x,y~\in~\mathfrak{g}.$$

For more information, see also e.g. this related Phys.SE post.

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$^1$ At the Lie group level, there can be topological obstructions.

Answer 3

This complexifying is useful only in the case of Lorentz algebra $so(1,3)$ (or $so(3,1)$). For a general Lie algebra, the complexifying unnecessarily complicates things for no apparent benefit. (it is understood that the assertion is in the context of representation of real world particles. If anyone can elaborate on the virtue of complexification of say, color algebra $su(3)_C$, I would be glad to know.)

The complexifying of Lorentz algebra all boils down to the identity $$ \gamma_0\gamma_1\gamma_2\gamma_3 \psi_{L} = - i\psi_{L},\\ \gamma_0\gamma_1\gamma_2\gamma_3 \psi_{R} = + i\psi_{R}, $$ where $\psi_{L}$ and $\psi_{R}$ are the left-handed spinor and right-handed spinor, respectively.

The reduction of the pseudoscalar $\gamma_0\gamma_1\gamma_2\gamma_3$ on left to the imaginary number $i$ on the right hand side of above identity is where the complexifying of Lorentz algebra is coming from.

Here is the full story:

The Lorentz algebra comprises 6 independent generators $$ \{\gamma_1\gamma_2, \gamma_2\gamma_3, \gamma_3\gamma_1,\gamma_0\gamma_3,\gamma_0\gamma_1,\gamma_0\gamma_2\}. $$

The mentioned identity $\gamma_0\gamma_1\gamma_2\gamma_3 \psi_{L/R} = \mp i\psi_{L/R}$ makes 3 of the generators degenerate for a given chirality. For example: $$ \gamma_0\gamma_1 \psi_{L/R} = -(\gamma_2\gamma_3)^2 (\gamma_0\gamma_1)\psi_{L/R}= -(\gamma_2\gamma_3) (\gamma_0\gamma_1\gamma_2\gamma_3)\psi_{L/R} = \pm i\gamma_2\gamma_3\psi_{L/R}, $$ (the first equality is resulted from $(\gamma_2\gamma_3)^2 = -1$) whereby real Lorentz boost ($\gamma_0\gamma_1$) is translated into imaginary rotation ($i\gamma_2\gamma_3$) and vice-versa: $$ e^{\theta\gamma_0\gamma_1} \psi_{L/R} = e^{\pm i\theta\gamma_2\gamma_3}\psi_{L/R}. $$

Therefore the original Lorentz algebra generators are turned into $$ \{\gamma_1\gamma_2, \gamma_2\gamma_3, \gamma_3\gamma_1,i\gamma_1\gamma_2, i\gamma_2\gamma_3, i\gamma_3\gamma_1\} $$ and $$ \{\gamma_1\gamma_2, \gamma_2\gamma_3, \gamma_3\gamma_1,-i\gamma_1\gamma_2, -i\gamma_2\gamma_3, -i\gamma_3\gamma_1\} $$ for left-handed spinor $\psi_{L}$ and right-handed spinor $\psi_{R}$, respectively.

Now you only have 3 independent (complexified) generators ($\{\gamma_1\gamma_2, \gamma_2\gamma_3, \gamma_3\gamma_1\} \sim su(2)$) for each chirality.

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Added note:

If one favors pure geometric algebra over matrix representation. One can replace the imaginary number $i$ with some even element of the geometric algebra acting on the right of a spinor, unlike the Lorentz transformation which is applied on the left of a spinor. One can thus live without complexification.

Answer 4

The Lie groups $SU(n)$ are all compact groups. Now the representation theory of any compact group (not necessarily smooth) is fully handled by the Peter-Weyl theorem.

Complexification is used in the analysis of the Lorentz group because of the isomorphisms:

$ltz(4)^{\mathbb{C}}\simeq 2su(2)^{\mathbb{C}} \simeq 2sl(2,\mathbb{C}) \simeq sl(2,\mathbb{C})^{\mathbb{C}}$

Furthermore, $Rep^{\mathbb{R}} g \simeq Rep^{\mathbb{C}} g^{\mathbb{C}}$. That is, there is a bijection between complex reps of $g$ and complex reps of the complexification of $g$. Note, the bijection is actually between isomorphism classes. This follows from the characterising/universal property of complexification.

Thus knowing all the reps of $su(2)$, via the Peter-Weyl theorem, allows us to construct all the complex reps of $ltz(4)$ and hence for $Ltz(4)$. However, this method does not generalise to higher (or lower) Lorentz groups. A more uniform method in that situation is given by looking at Clifford algebras.